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Chapter 3
Stoichiometry
Chapter 3
Chemical Stoichiometry
 Stoichiometry – The study of quantities of
materials consumed and produced in chemical
reactions.
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Section 3.1
Counting by Weighing



Objects behave as though they were all identical.
Atoms are too small to count.
Need average mass of the object.
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Section 3.1
Counting by Weighing
EXERCISE!
A pile of marbles weigh 394.80 g. 10 marbles weigh
37.60 g. How many marbles are in the pile?
Avg. Mass of 1 Marble =
37.60 g
= 3.76 g / marble
10 marbles
394.80 g = 105 marbles
3.76 g
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Section 3.2
Atomic Masses




12C
is the standard for atomic mass, with a mass of
exactly 12 atomic mass units (u).
The masses of all other atoms are given relative to this
standard.
Elements occur in nature as mixtures of isotopes.
Carbon = 98.89% 12C
1.11% 13C
< 0.01% 14C
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Section 3.2
Atomic Masses
Average Atomic Mass for Carbon
98.89% of 12 u + 1.11% of 13.0034 u =
exact number
(0.9889)(12 u) + (0.0111)(13.0034 u) =
12.01 u
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Section 3.2
Atomic Masses
Average Atomic Mass for Carbon


Even though natural carbon does not contain a single
atom with mass 12.01, for stoichiometric purposes, we
can consider carbon to be composed of only one type of
atom with a mass of 12.01.
This enables us to count atoms of natural carbon by
weighing a sample of carbon.
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Section 3.2
Atomic Masses
Schematic Diagram of a Mass Spectrometer
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Section 3.2
Atomic Masses
EXERCISE!
An element consists of 62.60% of an isotope with
mass 186.956 u and 37.40% of an isotope with mass
184.953 u.
 Calculate the average atomic mass and identify the
element.
186.2 u
Rhenium (Re)
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Section 3.3
The Mole



The number equal to the number of carbon atoms in
exactly 12 grams of pure 12C.
1 mole of something consists of 6.022 × 1023 units of
that substance (Avogadro’s number).
1 mole C = 6.022 × 1023 C atoms = 12.01 g C
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Section 3.3
The Mole
EXERCISE!
Calculate the number of iron atoms in a 4.48 mole
sample of iron.
2.70×1024 Fe atoms
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Section 3.4
Molar Mass

Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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Section 3.4
Molar Mass
CONCEPT CHECK!
Which of the following is closest to the average mass of
one atom of copper?
a)
b)
c)
d)
e)
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
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Section 3.4
Molar Mass
CONCEPT CHECK!
Calculate the number of copper atoms in a 63.55 g
sample of copper.
6.022×1023 Cu atoms
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Section 3.4
Molar Mass
CONCEPT CHECK!
Which of the following 100.0 g samples contains the
greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
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Section 3.4
Molar Mass
EXERCISE!
Rank the following according to number of atoms
(greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
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a)
c)
16
Section 3.4
Molar Mass
EXERCISE!
Consider separate 100.0 gram samples of each of the
following:
H2O, N2O, C3H6O2, CO2
 Rank them from greatest to least number of
oxygen atoms.
H2O, CO2, C3H6O2, N2O
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Section 3.5
Learning to Solve Problems
Conceptual Problem Solving

Where are we going?
 Read the problem and decide on the final goal.
 How do we get there?
 Work backwards from the final goal to decide where
to start.
 Reality check.
 Does my answer make sense? Is it reasonable?
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Section 3.6
Percent Composition of Compounds

Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound

For iron in iron(III) oxide, (Fe2O3):
mass % Fe =
2(55.85 g)
111.70 g
=
n 100% = 69.94%
2(55.85 g)+3(16.00 g) 159.70 g
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Section 3.6
Percent Composition of Compounds
EXERCISE!
Consider separate 100.0 gram samples of each of the
following:
H2O, N2O, C3H6O2, CO2
 Rank them from highest to lowest percent oxygen
by mass.
H2O, CO2, C3H6O2, N2O
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Section 3.7
Determining the Formula of a Compound
Formulas

Empirical formula = CH
 Simplest whole-number ratio
 Molecular formula = (empirical formula)n
[n = integer]
 Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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Section 3.7
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen

Device used to determine the mass percent of each
element in a compound.
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Section 3.7
Determining the Formula of a Compound
EXERCISE!
The composition of adipic acid is 49.3% C, 6.9% H,
and 43.8% O (by mass). The molar mass of the
compound is about 146 g/mol.
 What is the empirical formula?
C3H5O2
 What is the molecular formula?
C6H10O4
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Section 3.7
Determining the Formula of a Compound

A representation of a chemical reaction:
C2H5OH + 3O2
reactants

2CO2 + 3H2O
products
Reactants are only placed on the left side of
the arrow, products are only placed on the
right side of the arrow.
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Section 3.8
Chemical Equations
C2H5OH + 3O2



2CO2 + 3H2O
The equation is balanced.
All atoms present in the reactants are
accounted for in the products.
1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon dioxide
and 3 moles of water.
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Section 3.8
Chemical Equations


The balanced equation represents an overall ratio of
reactants and products, not what actually “happens”
during a reaction.
Use the coefficients in the balanced equation to decide
the amount of each reactant that is used, and the
amount of each product that is formed.
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Section 3.9
Balancing Chemical Equations
Writing and Balancing the Equation for a Chemical Reaction
1. Determine what reaction is occurring. What are the reactants,
the products, and the physical states involved?
2. Write the unbalanced equation that summarizes the reaction
described in step 1.
3. Balance the equation by inspection, starting with the most
complicated molecule(s). The same number of each type of atom
needs to appear on both reactant and product sides. Do NOT
change the formulas of any of the reactants or products.
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Section 3.9
Balancing Chemical Equations
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Section 3.9
Balancing Chemical Equations
EXERCISE!
Which of the following correctly balances the chemical
equation given below? There may be more than one correct
balanced equation. If a balanced equation is incorrect, explain
what is incorrect about it.
CaO + C
CaC2 + CO2
I.
II.
III.
IV.
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CaO2 + 3C
2CaO + 5C
CaO + (2.5)C
4CaO + 10C
CaC2 + CO2
2CaC2 + CO2
CaC2 + (0.5)CO2
4CaC2 + 2CO2
Section 3.9
Balancing Chemical Equations
CONCEPT CHECK!
Which of the following are true concerning balanced chemical
equations? There may be more than one true statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each substance you
have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the
substances used.
V. The sum of the coefficients on the reactant side equals the
sum of the coefficients on the product side.
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Section 3.9
Balancing Chemical Equations
Notice




The number of atoms of each type of element must be
the same on both sides of a balanced equation.
Subscripts must not be changed to balance an equation.
A balanced equation tells us the ratio of the number of
molecules which react and are produced in a chemical
reaction.
Coefficients can be fractions, although they are usually
given as lowest integer multiples.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Stoichiometric Calculations

Chemical equations can be used to relate the masses of
reacting chemicals.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant or product to
moles of that substance.
3. Use the balanced equation to set up the appropriate
mole ratios.
4. Use the appropriate mole ratios to calculate the
number of moles of the desired reactant or product.
5. Convert from moles back to grams if required by the
problem.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
EXERCISE!
Consider the following reaction:
P4(s) + 5 O2(g) ® 2 P2O5(s)
If 6.25 g of phosphorus is burned, what mass of oxygen does it
combine with?
8.07 g O2
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
EXERCISE!
(Part I)
Methane (CH4) reacts with the oxygen in the air to
produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air to
produce nitrogen monoxide and water.
 Write balanced equations for each of these
reactions.
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
EXERCISE!
(Part II)
Methane (CH4) reacts with the oxygen in the air to
produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the air to
produce nitrogen monoxide and water.
 What mass of ammonia would produce the same
amount of water as 1.00 g of methane reacting
with excess oxygen?
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Section 3.10
Stoichiometric Calculations:
Amounts of Reactants and Products
Let’s Think About It

Where are we going?
 To find the mass of ammonia that would produce the
same amount of water as 1.00 g of methane reacting
with excess oxygen.
 How do we get there?
 We need to know:
 How much water is produced from 1.00 g of methane and
excess oxygen.
 How much ammonia is needed to produce the amount of
water calculated above.
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Section 3.11
The Concept of Limiting Reactant
Limiting Reactants


Limiting reactant – the reactant that runs out first and
thus limits the amounts of products that can be formed.
Determine which reactant is limiting to calculate
correctly the amounts of products that will be formed.
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Section 3.11
The Concept of Limiting Reactant
Limiting Reactants
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Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Stoichiometric mixture
 N2(g) + 3H2(g)
2NH3(g)
Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Limiting reactant mixture
 N2(g) + 3H2(g)  2NH3(g)
Section 3.11
The Concept of Limiting Reactant
A. The Concept of Limiting Reactants
 Limiting reactant mixture
 N2(g) + 3H2(g)
2NH3(g)
 Limiting reactant is the reactant that runs out first.
 H2
Section 3.11
The Concept of Limiting Reactant
Limiting Reactants



The amount of products that can form is limited by the
methane.
Methane is the limiting reactant.
Water is in excess.
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Section 3.11
The Concept of Limiting Reactant
CONCEPT CHECK!
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2
a)
b)
c)
d)
e)
2H2O
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
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Section 3.11
The Concept of Limiting Reactant
Notice

We cannot simply add the total moles of all the
reactants to decide which reactant mixture makes the
most product. We must always think about how much
product can be formed by using what we are given, and
the ratio in the balanced equation.
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Section 3.11
The Concept of Limiting Reactant
CONCEPT CHECK!
 You know that chemical A reacts with chemical B.
You react 10.0 g of A with 10.0 g of B.
 What information do you need to know in order
to determine the mass of product that will be
produced?
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Section 3.11
The Concept of Limiting Reactant
Let’s Think About It

Where are we going?
 To determine the mass of product that will be
produced when you react 10.0 g of A with 10.0 g of
B.
 How do we get there?
 We need to know:
 The mole ratio between A, B, and the product they form.
In other words, we need to know the balanced reaction
equation.
 The molar masses of A, B, and the product they form.
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Section 3.11
The Concept of Limiting Reactant
EXERCISE!
You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that the
molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the balanced equation:
A + 2B 2C
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Section 3.11
The Concept of Limiting Reactant
Percent Yield

An important indicator of the efficiency of a particular
laboratory or industrial reaction.
Actual yield
 100%  percent yield
Theoretica l yield
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Section 3.11
The Concept of Limiting Reactant
EXERCISE!
Consider the following reaction:
P4(s) + 6F2(g)
4PF3(g)
 What mass of P4 is needed to produce 85.0 g of
PF3 if the reaction has a 64.9% yield?
46.1 g P4
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